Designing Control Systems - MIT OpenCourseWare...Designing a Control System Today’s goal: optimizing the design of a control system. Example: wallFinder System Using feedback to

6.01: Introduction to EECS I

Designing Control Systems

March 8, 2011

Midterm Examination #1

Time: Tonight, March 8, 7:30 pm to 9:30 pm

Location: Walker Memorial (if last name starts with A-M)

10-250 (if last name starts with N-Z)

Coverage: Everything up to and including Design Lab 5.

You may refer to any printed materials that you bring to exam.

You may use a calculator.

You may not use a computer, phone, or music player.

No software lab this week.

Signals and Systems

Multiple representations of systems, each with particular strengths.

Difference equations are mathematically compact.

y[n] = x[n] + p0 y[n − 1]

Block diagrams illustrate signal flow paths from input to output.

Operators use polynomials to represent signal flow compactly.

Delay

+

p0

X Y

Y = X + p0RY

System Functionals represent systems as operators.

Y 1 Y = HX ; H = =

X 1− p0R

Feedback, Cyclic Signal Paths, and Poles

The structure of feedback produces characteristic behaviors.

Feedback produces cyclic signal flow paths.

Delay

+X Y

Cyclic signal flow paths → persistent responses to transient inputs.

Delay

+

p0

X Y

We can characterize persistent responses (called modes) with poles.

−1 0 1 2 3 4n

y[n] = pno ; n ≥ 0

Designing a Control System

Today’s goal: optimizing the design of a control system.

Example: wallFinder System

Using feedback to control position (lab 4) can lead to bad behaviors.

di[n] = desiredFrontdo[n] = distanceFront

t

do

k = −0.5 t

do

k = −1

t

do

k = −2 t

do

k = −8

What causes these different types of responses ?

Is there a systematic way to optimize the gain k ?

( )

( )

Analysis of wallFinder System: Review

Response of system is concisely represented with difference equation.

proportional controller: v[n] = ke[n] = k di[n]− ds[n]

locomotion: do[n] = do[n − 1]− Tv[n − 1]

sensor with no delay: ds[n] = do[n]


The difference equations provide a concise description of behavior.

do[n] = do[n − 1]− Tv[n − 1] = do[n − 1]− Tk di[n − 1]− do[n − 1]

However it provides little insight into how to choose the gain k.

Analysis of wallFinder System: Block Diagram

A block diagram for this system reveals two feedback paths.


proportional controller: v[n] = ke[n] = k ( di[n]− ds[n]

) locomotion: do[n] = do[n − 1]− T v[n − 1]

sensor with no delay: ds[n] = do[n]

+ k −T + RDi Do−

V

Analysis of wallFinder System: System Functions

Simplify block diagram with R operator and system functions.

Start with accumulator.


What is the input/output relation for an accumulator?

+ RX YW

Y = RW = R(X + Y )

Y = R X 1−R

This is an example of a recurring pattern: Black’s equation.

Check Yourself

Determine the system function H = Y X

.

+ F

G

X Y

1. F

1− F G 2.

F 1 + F G

3. F + 1 1− G

4. F × 1

1− G

5. none of the above

+ F

G

X YW

Black’s Equation

Y Determine the system function H = .

X

Y = FW = F (X +GY ) = FX + F GY

Y F X ≡ H =

1− FG

forward gain F closed-loop gain H =

1− loop gain FG

Check Yourself

Determine the system function H = Y X

. 1

+ F

G

X YW

1. F

1− F G 2.

F 1 + F G

3. F + 1 1− G

4. F × 1

1− G


Black’s Equation

Black’s equation has two common forms.

+ F

G

X YW

+ F

G

X YW

−

H = F H = F

1 + F G 1− FG

Difference is equivalent to changing sign of G.

Right form is useful in most control applications where the goal is

to make Y converge to X.

Analyzing wallFinder: System Functions

Simplify block diagram with R operator and system functions.


Replace accumulator with equivalent block diagram.

+ k −T R1−R

Di Do−

Now apply Black’s equation a second time:

−kT RDo −kT R −kT R= 1−R = =

1− (1 + kT )RDi 1 + −kT R 1−R− kT R1−R


We can represent the entire system with a single system function.



+ k −T R1−R

Di Do−

Equivalent system with a single block:

−kTR1− (1 + kT )RDi Do

Modular! But we still need a way to choose k.

Analyzing wallFinder: Poles

The system function contains a single pole at z = 1 + kT .

Do −kT R= Di 1− (1 + kT )R

The numerator is just a gain and a delay.

The whole system is equivalent to the following:

R 1−p0 +

p0 R

Di Do

where po = 1 + kT . Here is the unit-sample response for kT = −0.2:

0n

h[n]

0.2

Analyzing wallFinder

We are often interested in the step response of a control system.

Start the output do[n] at zero while the input is held constant at one.


Step Response

Calculating the unit-step response.

Unit-step response s[n] is response of H to the unit-step signal u[n],which is constructed by accumulation of the unit-sample signal δ[n].

+

R

Hδ[n] u[n] s[n]

Commute and relabel signals.

+

R

Hδ[n] h[n] s[n]

The unit-step response s[n] is equal to the accumulated responses

to the unit-sample response h[n].


The step response of the wallFinder system is slow because the

unit-sample response is slow.

0n

h[n]

0.2

1

0n

s[n]


The step response is faster if kT = −0.8 (i.e., p0 = 0.2).

0.8

0n

h[n]

0n

s[n]

1


The poles of the system function provide insight for choosing k.

Do −kT R (1− po)R Di

=1− (1 + kT )R

=1− poR

; p0 = 1 + kT

1 Re z

Im z

0 < p0 < 1−1 < kT < 0

monotonicconverging

1 Re z

Im z

−1 < p0 < 0−2 < kT < −1

alternating

converging

1 Re z

Im z

p0 < −1kT < −2

alternating

diverging

Check Yourself

Find kT for fastest convergence of unit-sample response.

Do Di

= −kT R1− (1 + kT )R

1. kT = −2 2. kT = −1 3. kT = 0 4. kT = 1 5. kT = 2 0. none of the above

Check Yourself

Find kT for fastest convergence of unit-sample response.

Do −kT R= Di 1− (1 + kT )R

If kT = −1 then the pole is at z = 0.

Do = −kT R Di 1− (1 + kT )R

= R

unit-sample response has a single non-zero output sample, at n = 1.

Check Yourself

Find kT for fastest convergence of unit-sample response. 2

Do Di

= −kT R1− (1 + kT )R

1. kT = −2 2. kT = −1 3. kT = 0 4. kT = 1 5. kT = 2 0. none of the above

( ) ( )


The optimum gain k moves robot to desired position in one step.

kT = −1

di[n] = desiredFront=1m

do[n] = distanceFront=2m

1 1k = − = −

1/10 = −10

T

v[n] = k di[n]− do[n] = −10 1− 2 = 10 m/s

exactly the right speed to get there in one step!

Analyzing wallFinder: Space-Time Diagram



position

time




position

time

v = 10




position

time

v = 10




position

time

v = 10v = 0




position

time

v = 10v = 0




position

time

v = 10v = 0v = 0




position

time

v = 10v = 0v = 0v = 0v = 0v = 0v = 0

( )

Analysis of wallFinder System: Adding Sensory Delay

Adding delay tends to destabilize control systems.

proportional controller: v[n] = ke[n] = k di[n]− ds[n]

locomotion: do[n] = do[n − 1]− Tv[n − 1]

sensor with delay: ds[n] = do[n − 1]





position

time




position

time

v = 10




position

time

v = 10




position

time

v = 10v = 0




position

time

v = 10v = 0




position

time

v = 10v = 0v = −10




position

time

v = 10v = 0v = −10




position

time

v = 10v = 0v = −10v = −10




position

time

v = 10v = 0v = −10v = −10




position

time

v = 10v = 0v = −10v = −10v = 0




position

time

v = 10v = 0v = −10v = −10v = 0

Analysis of wallFinder System: Block Diagram

Incorporating sensor delay in block diagram.


proportional controller: v[n] = ke[n] = k ( di[n]− ds[n]

) locomotion: do[n] = do[n − 1]− T v[n − 1]

sensor with delay: ds[n] = do[n − 1]

+ k −T + R

R

Di Do−

V

+ k −T + R

R

Di Do−

V


We can represent the entire system with a single system function.

Check Yourself

+ k −T + R

R

Di Do−

V

Find the system function H = Do Di

.

1. kT R1− R

2. −kT R

1 +R − kT R2

3. kT R1− R

− kT R 4. −kT R

1− R − kT R2


Check Yourself


.

+ k −T + R

R

Di Do−

V


+ k −T R1−R

R

Di Do−

−kT RDo = 1−R

2 = −kT R 2Di 1 + −kT R

1−R− kT R

1−R

Check Yourself

+ k −T + R

R

Di Do−

V


. 4

1. kT R1− R

2. −kT R

1 +R − kT R2

3. kT R1− R

− kT R 4. −kT R

1− R − kT R2


√


1 Substitute for R in the system functional to find the poles.

z

Do −kT R −kT 1 −kT z = = z = Di 1−R− kT R2 1− z

1 − kT z1 2 z2 − z − kT

The poles are then the roots of the denominator.

1 (

1)2

z = + kT 2±

2

Poles

Poles can be identified by expanding the system functional in partial

fractions.

Y b0 + b1R + b2R2 + b3R3 += · · · X 1 + a1R + a2R2 + a3R3 + · · ·

Factor denominator:

Y b0 + b1R + b2R2 + b3R3 += · · · X (1− p0R)(1 − p1R)(1 − p2R)(1 − p3R) · · ·

Partial fractions:

Y e0 e1 e2 X

=1− p0R

+1− p1R

+1− p2R

+ · · · + f0 + f1R + f2R2 + · · ·

The poles are pi for 0 ≤ i < n where n is the order of the denominator.

One geometric mode pin arises from each factor of the denominator.

√( ( )Feedback and Control: Poles

If kT is small, the poles are at z ≈ −kT and z ≈ 1 + kT .

z =)2+ kT = 1

2 1± √

1 + 4kT ≈ (1± (1 + 2kT )) = 1 + kT, −kT 12

1 Re z

Im zz-planekT ≈ 0

Pole near 0 generates fast response.

Pole near 1 generates slow response.

Slow mode (pole near 1) dominates the response.

12

12±

Feedback and Control: Poles

As kT becomes more negative, the poles move toward each other

and collide at z = .12 when kT = −14

z = 12 ± √(1

2 )2 + kT = 1

2 ± √(1

2 )2 − 1

4 =12 ,

12

21 Re z

Im zz-plane

kT = −14

Persistent responses decay. The system is stable.

√ √


If kT < −1/4, the poles are complex.

z = 12 ±

(12 )2 + kT = 1

2 ± j −kT − (1

2 )2

1 Re z

Im zz-planekT = −1

Complex poles → oscillations.

Same oscillation we saw earlier!



position

time

v = 10v = 0v = −10v = −10v = 0

Check Yourself

1 Re z


What is the period of the oscillation?

1. 1 2. 2 3. 3

4. 4 5. 6 0. none of above

Check Yourself

1 p0 = 2

± j

1 Re z


√3

2 = e±jπ/3

n = e±jπn/3 p0

e±j0π/3 , e±jπ/3 , e±j2π/3 , e±j3π/3 , e±j4π/3 , e±j5π/3 , e±j6π/3 ︸︷︷︸︸︷︷︸ 1 e±j2π=1

Check Yourself

1 Re z


What is the period of the oscillation? 5

1. 1 2. 2 3. 3

4. 4 5. 6 0. none of above


The closed-loop poles depend on the gain.

1 Re z

Im zz-plane

If kT : 0→ −∞: then z1, z2 : 0, 1→ 1 2 ,

1 2 → 1

2 ± j∞

Check Yourself

1 Re z

Im zz-plane

closed-loop poles

12±

√(12

)2+ kT

Find kT for fastest response.

1. 0 2. −14 3. −1

2 4. −1 5. −∞ 0. none of above

Check Yourself

z = 1 2±

√(1 2

)2 + kT

1 The dominant pole always has a magnitude that is ≥

2.

1 It is smallest when there is a double pole at z =

2.

1 Therefore, kT = −

4.

Check Yourself

1 Re z

Im zz-plane

closed-loop poles

12±

√(12

)2+ kT

Find kT for fastest response. 2

1. 0 2. −14 3. −1

2 4. −1 5. −∞ 0. none of above

1 Re z

Im z

1 Re z

Im z

Destabilizing Effect of Delay

Adding delay in the feedback loop makes it more difficult to stabilize.

Ideal sensor: ds[n] = do[n]

More realistic sensor (with delay): ds[n] = do[n − 1]

Fastest response without delay: single pole at z = 0. 1

Fastest response with delay: double pole at z =2. much slower!

Destabilizing Effect of Delay

Adding more delay in the feedback loop is even worse.

More realistic sensor (with delay): ds[n] = do[n − 1]

Even more delay: ds[n] = do[n − 2]

1 Re z

Im z

21 Re z

Im z

Fastest response with delay: double pole at z = 2.

1

Fastest response with more delay: double pole at z = 0.682.

even slower →

Check Yourself

R R R+X Y

How many of the following statements are true?

1. This system has 3 poles.

2. unit-sample response is the sum of 3 geometric sequences.

3. Unit-sample response is y[n] : 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . .


5. One of the poles is at z = 1.

Check Yourself

R R R+X Y

How many of the following statements are true? 4

1. This system has 3 poles.

2. unit-sample response is the sum of 3 geometric sequences.



5. One of the poles is at z = 1.

Designing Control Systems: Summary

System Functions provide a convenient summary of information that

is important for designing control systems.

The long-term response of a system is determined by its dominant

pole — i.e., the pole with the largest magnitude.

A system is unstable if the magnitude of its dominant pole is > 1.

A system is stable if the magnitude of its dominant pole is < 1.

Delays tend to decrease the stability of a feedback system.

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Designing Control Systems - MIT OpenCourseWare...Designing a Control System Today’s goal: optimizing the design of a control system. Example: wallFinder System Using feedback to